Suppose $T \in \mathcal{L}(V)$. Prove that $T$ is invertible if and only if there exists a unique isometry $S \in \mathcal{L}(V)$ such that $T = S \sqrt{T^* T}$.
I have already proved the “if” part. But I have nothing for the only if part. Some proof?
show that the polar part $P$ is unique, thereafter the notation$\sqrt{T^\ast T}$ makes sense. This part of the uniquenss: the squared root $R$ of a semipositive definite operator $N$ is unique, can be shown directly using the spectral decomposition of $R$. Let $R=\sum_{i=1}^{k} r_i v_i v_i^\ast$ be the spectral decomposition of $R$, where $r_i>0$ are the singular values of $R$, then $N=R^2=\sum_{i=1}^{k} r_i^2 v_i v_i^\ast$ is the spectral decomposition of $N$, so the $R$ is unique, even though the eigenvectors $v_i$ are not unique.
Suppose that $T = SP$ is not invertible, then $P$ is also not invertible. Now one can replace $S$ by $SS'$, where $S'$ is an isometry such that $S'v = v$ for all eigenvectors $v$ of non-zero eigenvalue. Hence, $S$ is not unique.
On the other hand, if $T$ is invertible, $S = TP^{-1}$, hence $S$ is unique.